Cartesian Coordinates Plane Strain

16 If the deformation of a cylindrical body is such that there is no axial components of the displacement and that the other components do not depend on the axial coordinate, then the body is said to be in a state of plane strain. If e3 is the direction corresponding to the cylindrical axis, then we have

and the strain components corresponding to those displacements are

Ox-2

1 (dui du2

and the non-zero stress components are Tii, Ti2, T22, T33 where

17 Considering a static stress field with no body forces, the equilibrium equations reduce to:

dTii + dTi2

dxi dx2 dTi2 + 8X22 dxi dx2 dTss dxi

we note that since T33 = T33(x\, x2), the last equation is always satisfied.

is Hence, it can be easily verified that for any arbitrary scalar variable $, if we compute the stress components from

d2$

(10.17)

T11=

dx2

d2$

(10.18)

T22 =

dx2

d 2$

(10.19)

T12 =

dxidx2

then the first two equations of equilibrium are automatically satisfied. This function $ is called Airy stress function.

19 However, if stress components determined this way are statically admissible (i.e. they satisfy equilibrium), they are not necessarily kinematically admissible (i.e. satisfy compatibility equations).

20 To ensure compatibility of the strain components, we obtain the strains components in terms of $ from Hooke's law, Eq. 5.1 and Eq. 10.15.

Eii E22 Ei2

dx2 j

21 For plane strain problems, the only compatibility equation, 4.140, that is not automatically satisfied is d2Eu d2E22 n d2 E12

dxidx2

thus we obtain the following equation governing the scalar function $

or d4$ d4 $ d4$ T7 + 2jn^-2 + TT =0 or V4$ = 0 dxi dxfdx2 dx4

Hence, any function which satisfies the preceding equation will satisfy both equilibrium and kinematic and is thus an acceptable elasticity solution.

22 We can also obtain from the Hooke's law, the compatibility equation 10.21, and the equilibrium equations the following

23 Any polynomial of degree three or less in x and y satisfies the biharmonic equation (Eq. 10.23). A systematic way of selecting coefficients begins with

24 The stresses will be given by

oo oo

J2J2n(n - 1)CmnXmyn-2

25 Substituting into Eq. 10.23 and regrouping we obtain oo oo

J2 J2[(m+2)(m+1)m(m-1)Cm+2,u-2+2m(m-1)n(n-1)Cmn+(n+2)(n+1)n(n-1)Cm-2,u+2]xm-2yn-2 = 0

but since the equation must be identically satisfied for all x and y, the term in bracket must be equal to zero.

(m+2)(m +1)rn(rn-1)Cm+2,n-2 +2m(m-1)n(n-1)Cmn + (n+2)(n +1)n(n-1)Cm_2,n+2 = 0 (10.28)

Hence, the recursion relation establishes relationships among groups of three alternate coefficients which can be selected from

0

0

C02

C03

C04

C05 C06 • • •

0

C11

C12

C13

C14

C15 • • •

C20

C21

C22

C23

C24

C30

C31

C32

C33

(10.29)

C40

C41

C42

C50

C51

C60

(4)(3)(2)(1)C4o + (2)(2)(1)(2)(1)C22 + (4)(3)(2)(1)Co4 = 0

10.2.1.1 Example: Cantilever Beam

26 We consider the homogeneous fourth-degree polynomial

$4 = C40X4 + C31x3y + C22x2y2 + C\3xy3 + C04 y4 with 3C40 + C22 + 3Co4 = 0,

27 The stresses are obtained from Eq. 10.26-a-10.26-c

Txx = 2C22X2 +6Ci3xy + 12Co4 y2 Tyy = I2C40X2 +6C3ixy + 2C22 y2

These can be used for the end-loaded cantilever beam with width b along the z axis, depth 2a and length L.

28 If all coefficients except C13 are taken to be zero, then

29 This will give a parabolic shear traction on the loaded end (correct), but also a uniform shear traction Txy = —3C13a2 on top and bottom. These can be removed by superposing uniform shear stress Txy = +3C13a2 corresponding to $2 = —3C13a2xy. Thus

note that C20 = C02 = 0, and C11 = —3C13a2. 30 The constant C13 is determined by requiring that hence and the solution is

T Tx

4a3b

3P P

bxy 3P

4abxy 4a3b xy

T xy

2a3b

31 We observe that the second moment of area for the rectangular cross section is I = 5(2a)3/12 = 2a3b/3, hence this solution agrees with the elementary beam theory solution

Cllxy H Cl3xy

4abxy 4a3b xy

10.2.2 Polar Coordinates 1G.2.2.1 Plane Strain Formulation

32 In polar coordinates, the strain components in plane strain are, Eq. 9.46

and the equations of equilibrium are

33 Again, it can be easily verified that the equations of equilibrium are identically satisfied if

+ r2 dO2

(10.41)

Tee =

d2$ dr2

(10.42)

Tre =

d dr

( 1 d$ \ \r~30)

(10.43)

34 In order to satisfy the compatibility conditions, the cartesian stress components must also satisfy Eq. 10.24. To derive the equivalent expression in cylindrical coordinates, we note that Tn + T22 is the first scalar invariant of the stress tensor, therefore

r dr r2 dO2 or2

35 We also note that in cylindrical coordinates, the Laplacian operator takes the following form

dr2 r dr r2 BO2

Thus, the function $ must satisfy the biharmonic equation

10.2.2.2 Axially Symmetric Case

37 If $ is a function of r only, we have

r dr and

dr4 r dr3 r2 dr2 r3 dr 38 The general solution to this problem; using Mathematical

DSolve[phi''''[r]+2 phi'''[r]/r-phi''[r]/r"2+phi'[r]/r"3==0,phi[r],r]

39 The corresponding stress field is

and the strain components are (from Sect. 9.8.1)

duT dr

+ (1 - 3v - 4v2)B + 2(1 - v - 2v2)Blnr + 2(1 - v - 2v2)C

+ (3 - v - 4v2)B + 2(1 - v - 2v2)B lnr + 2(1 - v - 2v2)C

4r0B

(1 + v)A - (1 + v)Br + 2(1 - v - 2v2)r ln rB + 2(1 - v - 2v2)rC

0 0

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